Free Energy and Equilibrium
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AP Chemistry › Free Energy and Equilibrium
A student observes the reaction $\mathrm{2HF(aq) \rightleftharpoons H_2F_2(aq)}$ in solution at constant temperature. At one moment, $\Delta G<0$ for the reaction as written. Which statement best describes the system’s position relative to equilibrium?
The system is at equilibrium because $\Delta G<0$ indicates the minimum free energy state is reached.
The system will shift toward products until $\Delta G \approx 0$.
The system must contain only $\mathrm{H_2F_2}$ because negative $\Delta G$ implies completion.
The system cannot reach equilibrium because $\Delta G<0$ means the reverse reaction cannot occur.
The system will shift toward reactants until $\Delta G \approx 0$.
Explanation
This question tests determining a system's equilibrium position from ΔG's sign. For 2HF(aq) ⇌ H₂F₂(aq), ΔG < 0 means the system is not at equilibrium and will shift toward products (H₂F₂) until ΔG ≈ 0. This indicates current concentrations make Q < K, favoring dimer formation. Choice B best describes this. Choice C tempts but is incorrect, suggesting shift to reactants, due to the misconception of negative ΔG favoring reverse. In aqueous equilibria, use ΔG's sign to predict net shifts toward the spontaneous direction.
For $\mathrm{C(s) + CO_2(g) \rightleftharpoons 2CO(g)}$ at constant temperature, the system is at equilibrium and thus $\Delta G \approx 0$. Which statement is true about the reaction at this point?
Only reactants are present because $\Delta G \approx 0$ indicates no reaction can proceed.
Both the forward and reverse reactions are occurring, but there is no net change in composition.
The reverse reaction has stopped completely because $\Delta G \approx 0$ means products are maximum.
The forward reaction has stopped completely because $\Delta G \approx 0$ means no collisions occur.
Only products are present because $\Delta G \approx 0$ indicates completion.
Explanation
This question tests understanding that at equilibrium (ΔG ≈ 0), reactions continue microscopically without net change. For C(s) + CO₂(g) ⇌ 2CO(g), ΔG ≈ 0 means both forward and reverse reactions occur, but rates are equal, resulting in no net composition change. This dynamic equilibrium maintains constant concentrations despite ongoing reactions. Choice A is true about the system. Choice D tempts but is incorrect, assuming ΔG ≈ 0 means only products, due to the misconception that equilibrium implies completion. To distinguish, remember equilibrium involves equal rates, not cessation, and verify with ΔG = 0.
For $\mathrm{2CO(g) + O_2(g) \rightleftharpoons 2CO_2(g)}$ at constant temperature, the system is prepared so that $\Delta G$ for the forward reaction is positive. Which statement best describes the equilibrium tendency from that moment?
No net change will occur because a positive $\Delta G$ means the reaction is too slow.
The system is already at equilibrium because $\Delta G$ can be positive at equilibrium.
The system will produce only products because combustion reactions always go to completion.
The system will shift toward products until $\Delta G$ becomes more positive.
The system will shift toward reactants until $\Delta G$ approaches zero.
Explanation
This question evaluates the interpretation of positive ΔG in predicting the path to equilibrium. For 2CO(g) + O₂(g) ⇌ 2CO₂(g), ΔG > 0 for the forward reaction means the reverse is spontaneous, so the system shifts toward reactants until ΔG approaches zero at equilibrium. This is because positive ΔG indicates Q > K, favoring reactant formation. Choice A best describes this tendency. Choice B is a tempting distractor but incorrect, as it suggests shifting toward products, arising from the misconception that positive ΔG favors the forward direction. To solve these, use the rule that the spontaneous direction opposes the sign of ΔG for the written reaction.
A system at constant temperature is at equilibrium for $\mathrm{Br_2(l) \rightleftharpoons Br_2(aq)}$, so $\Delta G \approx 0$. Which statement best describes what $\Delta G \approx 0$ means here?
Precipitation is spontaneous and will proceed until all aqueous bromine is gone.
The process is impossible because $\Delta G \approx 0$ means the reaction cannot occur.
The equilibrium mixture must contain equal amounts of $\mathrm{Br_2(l)}$ and $\mathrm{Br_2(aq)}$.
Dissolution is spontaneous and will proceed until all liquid bromine is gone.
There is no net driving force for dissolution or precipitation under the current conditions.
Explanation
This question examines interpreting ΔG ≈ 0 in phase or solubility equilibria. For Br₂(l) ⇌ Br₂(aq), ΔG ≈ 0 at equilibrium means no net driving force for dissolution or precipitation, as the system is balanced. This indicates saturation where rates of dissolving and precipitating are equal. Choice A best describes this. Choice E tempts but is incorrect, assuming equal amounts, from the misconception that equilibrium requires equal quantities rather than equal rates. For solubility problems, use ΔG = 0 to identify saturation points without assuming equal concentrations.
A closed container at constant temperature contains the system $\mathrm{2NO_2(g) \rightleftharpoons N_2O_4(g)}$. At a certain moment, $\Delta G$ for the reaction as written is negative. Which statement best describes the relationship between $\Delta G$ and equilibrium?
Because $\Delta G < 0$, the reaction must go to completion with only $\mathrm{N_2O_4}$ present.
Because $\Delta G < 0$, the system is at equilibrium and will not change composition.
Because $\Delta G < 0$, the reverse reaction is spontaneous until $\Delta G$ becomes positive.
Because $\Delta G < 0$, the equilibrium constant must be less than 1 at this temperature.
Because $\Delta G < 0$, the system will shift toward $\mathrm{N_2O_4}$ until $\Delta G = 0$.
Explanation
This question tests understanding of spontaneous direction when ΔG < 0. When ΔG is negative for the reaction as written (2NO₂ → N₂O₄), the forward reaction is spontaneous, meaning the system will shift toward products (N₂O₄) until equilibrium is reached. The negative ΔG indicates the system can lower its free energy by forming more N₂O₄, and this process continues until ΔG = 0 at equilibrium. The system is not currently at equilibrium because ΔG ≠ 0. The misconception in choice A is that negative ΔG indicates equilibrium, when it actually indicates the forward reaction is favored. When ΔG < 0, the system spontaneously proceeds in the forward direction until reaching equilibrium.
For the reaction $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ at a constant temperature, a student measures the system and finds $\Delta G \approx 0$ for the reaction mixture. Which statement best describes the state of the system?
The system is not at equilibrium, because $\Delta G$ must be negative at equilibrium.
The system is at equilibrium only if the enthalpy change $\Delta H$ is also approximately zero.
The system is reactant-favored, so the reverse reaction must be faster than the forward reaction.
The system is at equilibrium, so the forward and reverse reaction rates are equal.
The system is product-favored, so the reaction will proceed forward until completion.
Explanation
This question tests understanding of the relationship between Gibbs free energy and equilibrium. When ΔG ≈ 0 for a reaction mixture, the system is at equilibrium, meaning the forward and reverse reaction rates are equal and there is no net change in concentrations. At equilibrium, the system has reached its minimum free energy state, and neither the forward nor reverse reaction is thermodynamically favored. The misconception in choice D is that ΔG must be negative at equilibrium, when in fact ΔG = 0 defines the equilibrium condition. To determine if a system is at equilibrium, check if ΔG = 0 for the current mixture composition.
For $\mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}$ at a fixed temperature, a reaction mixture is adjusted and then found to have $\Delta G \approx 0$. Which statement best connects $\Delta G$ to the equilibrium position?
$\Delta G \approx 0$ means products and reactants must be present in equal amounts.
$\Delta G \approx 0$ means the mixture is at equilibrium for its current composition.
$\Delta G \approx 0$ means the reaction cannot proceed in either direction under any conditions.
$\Delta G \approx 0$ means the equilibrium constant $K$ is approximately 0.
$\Delta G \approx 0$ means the reaction is strongly product-favored and will proceed forward rapidly.
Explanation
This question tests understanding of the meaning of ΔG ≈ 0 for a reaction mixture. When ΔG ≈ 0, the system is at equilibrium for its current composition, meaning there is no driving force for net change in either direction. At this point, the forward and reverse reactions occur at equal rates, and the reaction quotient Q equals the equilibrium constant K. The value of ΔG depends on the current mixture composition, not just the balanced equation. The misconception in choice B is confusing ΔG = 0 with K = 0, when actually ΔG = 0 occurs when Q = K regardless of K's value. To identify equilibrium, check if ΔG = 0 for the specific mixture composition.
For $\mathrm{X(aq) + Y(aq) \rightleftharpoons Z(aq)}$ at constant temperature, a student says: “If $\Delta G$ is negative, the reaction must already be at equilibrium.” Which statement best addresses this idea?
Incorrect, because $\Delta G<0$ indicates the reverse reaction is spontaneous until $\Delta G \approx 0$.
Incorrect, because $\Delta G<0$ indicates a net tendency to form products until $\Delta G \approx 0$.
Incorrect, because $\Delta G<0$ means the reaction cannot be reversible.
Correct, because equilibrium occurs whenever $\Delta G$ is negative.
Correct, because negative $\Delta G$ means the forward and reverse rates are equal.
Explanation
This question probes evaluating claims about ΔG and equilibrium states. For X(aq) + Y(aq) ⇌ Z(aq), the student's idea that negative ΔG means the reaction is at equilibrium is incorrect because ΔG < 0 indicates spontaneity toward products until ΔG ≈ 0 at equilibrium. Negative ΔG shows the system is not yet at minimum free energy. Choice C best addresses this. Choice D tempts but is wrong, as it flips to reverse spontaneity, due to the misconception of sign reversal. To assess such claims, contrast ΔG's role in non-equilibrium (spontaneity) versus equilibrium (zero) conditions.
For $\mathrm{H_2(g) + Cl_2(g) \rightleftharpoons 2HCl(g)}$ at a given temperature, the system is at equilibrium so $\Delta G \approx 0$. Which statement correctly connects $\Delta G$ to the equilibrium state?
$\Delta G \approx 0$ means the forward reaction is spontaneous and the reverse is not.
$\Delta G \approx 0$ means products are present in a greater amount than reactants.
$\Delta G \approx 0$ means the forward and reverse reaction rates are equal.
$\Delta G \approx 0$ means the reaction cannot proceed in either direction.
$\Delta G \approx 0$ means the equilibrium constant $K$ must equal 1 for this reaction.
Explanation
This question assesses connecting ΔG ≈ 0 to microscopic aspects of equilibrium. For H₂(g) + Cl₂(g) ⇌ 2HCl(g), ΔG ≈ 0 at equilibrium means forward and reverse rates are equal, maintaining constant concentrations. This dynamic state persists despite ongoing reactions. Choice C correctly connects this. Choice A is a tempting distractor but wrong, as it assumes K=1, from the misconception that ΔG=0 implies Q=1 specifically. Relate ΔG=0 to rate equality, not K's value, for accurate equilibrium descriptions.
For the reaction $\mathrm{A(g) \rightleftharpoons B(g)}$ at constant temperature, a student measures $\Delta G \approx 0$ for the forward direction. Which statement is consistent with this measurement?
The system is at equilibrium, so there is no net change in the amounts of $\mathrm{A}$ and $\mathrm{B}$.
The system is reactant-favored, so $\mathrm{A}$ must be present at a higher concentration than $\mathrm{B}$.
The reverse reaction is spontaneous, so $\mathrm{B}$ will be completely converted to $\mathrm{A}$.
The forward reaction is spontaneous, so $\mathrm{A}$ will be completely converted to $\mathrm{B}$.
The system is product-favored, so $\mathrm{B}$ must be present at a higher concentration than $\mathrm{A}$.
Explanation
This question examines how ΔG ≈ 0 indicates a system at equilibrium with no net composition change. For A(g) ⇌ B(g), ΔG ≈ 0 for the forward direction means the system is at equilibrium, with no net change in amounts of A and B as rates are equal. At this point, the free energy is at a minimum, preventing spontaneous conversion. Choice A is consistent with this measurement. Choice B tempts by assuming product-favorability implies higher B concentration, but this is incorrect as ΔG = 0 doesn't specify favorability, stemming from mixing ΔG with ΔG°. To interpret such data, recall that ΔG = 0 solely denotes equilibrium, independent of K's value.